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If all of the telephone extensions in a certain company must [#permalink]
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22 Jan 2005, 14:03
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If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of fourdigit extensions that the company can have? (A) 4 (B) 6 (C) 12 (D) 16 (E) 24
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If all of the telephone extensions in a certain company must [#permalink]
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22 Jan 2005, 14:16
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let the 4 digits be
the possible ways are
A B C D
1 * 2 *3 * 2 = 12
total 12 ways
the logic is simple
4 digits are there so the units shud be 2 or 6 tht can be counted in 2 ways
for the next digit there are 3 left so 3 ways the next digit 2 left 2 ways last one can be chosen in 1 way for the lefot over digit
so 1* 2* 3* 2 = 12
P.S. but also if repetetions are allowed then it will be
4* 4 * 4* 2 = 128



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Re: If all of the telephone extensions in a certain company must [#permalink]
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23 Jan 2005, 17:27
Why is it not 48? (I know it's not an option)
Like 4*3*2*2?
Rationale:
We are combining 4 digit extensions which must be even numbers. ABCD i.e., D but be either 2 or 6 from the given digits 1, 2, 3, 6.
Now we are left with the first 3 spaces which can be filled with any of these 4 given digits 1, 2, 3 or 6 which implies 4*3*2.
I stand corrected though.



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Re: If all of the telephone extensions in a certain company must [#permalink]
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24 Jan 2005, 00:44
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Folaa3 wrote: Now we are left with the first 3 spaces which can be filled with any of these 4 given digits 1, 2, 3 or 6 which implies 4*3*2.
Each number must use all of the four numbers. So after you pick the unit digit, you only have three numbers to pick, not four.



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Re: If all of the telephone extensions in a certain company must [#permalink]
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24 Jan 2005, 09:39
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nice question,
*** 1st, the number can be expressed as "ABC2" (ending in 2)
so, there are 3*2*1*1 = 6 ways to express the number using 1,2,3,6
*** 2nd, the number can be expressed as "ABC6" (ending in 6)
so, there are 3*2*1*1 = 6 more ways to express the number using 1,2,3,6
therefore: 6 + 6 = 12



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Re: PS telephones numbers [#permalink]
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09 Jun 2005, 02:47
mandy wrote: :) .hello Does anyone knows how to approach this one thanks PS If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of fourdigit extensions that the company can have? (A) 4 (B) 6 (C) 12 (D) 16 (E) 24
The four digits available are 1,2,3 and 6
as the extensions can only be even numbers, the last digit has to be either 2 or 6
If it is 2...it leaves 1,3 and 6 for the first three digits...these can be arranged in 3! ways
If the last digit is 6...it leaves 1,2 and 3 for the first three digits...these can be arranged in 3! ways
Thus answer is 2*3! = 12



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Re: PS telephones numbers [#permalink]
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09 Jun 2005, 09:24
mandy wrote: :) .hello Does anyone knows how to approach this one thanks PS If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of fourdigit extensions that the company can have? (A) 4 (B) 6 (C) 12 (D) 16 (E) 24
4!/2 ,
The total no. extensions possible is 4!.
Of which, equal no. extns can be odd and even.
Hence the ones which are even is 4!/2.
HMTG.



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Re: PS  1000PS [#permalink]
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20 Apr 2007, 22:40
[quote="surbab"]If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of fourdigit extensions that the company can have?
(A) 4
(B) 6
(C) 12
(D) 16
(E) 24
Please explain the answer.[/quote]
(C)
Let's see... the conditions are that:
 4digits telephone numbers
 the only digits we can consider are 1,2,3,6
 the telephone extensions must be even numbers, so they will finish in 2 or 6
 ALL the digits (1,2,3,6) should be used in the same telephone number
total # of even numbers = total # with final digit 2 + total # with final digit 6
= 3*2*1 (the 4th digit is nr. 2) + 3*2*1 (the 4th digit is nr. 6)
=12
The easiest way to solve this problem is to think the telephone number digit by digit.
e.g.
4th digit: it must be even, so lets assume that it finishes in 2 > 1
1st digit: we have 3 options (1 or 3 or 6) > 3
2nd digit: we have 2 options (1,3 or 1,6 or 3,6) > 2
3rd digit: we have only 1 nr left (1 or 3 or 6) > 1
When you multiply all the numbers calculated per digit you get the number of possibilities: 1*3*2*1 = 6



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Since the phone number must be even, the unit's digit can be either 2 or 6.
When the unit's digit is 2 > number of possibilities is 3! = 6
When the unit's digit is 6 > number of possibilities is 3! = 6
Largest number of extensions = 6 + 6 = 12
Answer: C



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For the extension to be even the last digit should be 2 or 6.
Number of combinations with 2 be the last digit = 6
Number of combinations with 6 be the last digit = 6
Total possible cases = 12
Hence answer is 'C'



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If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of fourdigit extensions that the company can have?
(A) 4
(B) 6
(C) 12
(D) 16
(E) 24
Please explain the answer.
All possibilities for the combination of 4 digits is 4!=4x3x2x1= 24. Since we only need even numbers an we do have 2 even and 2 odd numbers it is 50% or 12!
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Re: telephone extensions [#permalink]
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08 Jun 2009, 13:24
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it should be 12 
digit 1 * digit 2 * digit 3 * digit 4
digit 4  last digit is having 2 option only even. digit 3  left out 3 digit 2 has left out 2 digit 1 has left out 1.
multiply all ways u get 1*2*3*2 = 12....
please confirm.



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Re: telephone extensions [#permalink]
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09 Jun 2009, 16:04
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The answer is C.
You are on the right track..the total number of permutations is 4! = 24
However there is a restriction that all extensions should be even i.e. the last digit of each extension should be an even number i.e. the last digit should be either 2 or 6. As there are 2 even digits and two odd digits. Half of the permutations should be even and the remaining odd.
Divide 24/2 results into 12



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Re: telephone extensions [#permalink]
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09 Jun 2009, 16:29
one more way.... let digit4 = 6, the 3! is the way to arrange rest 3 digit, so total 6 ways let digit4 = 2, the 3! is the way to arrange rest 3 digit, so total 6 ways
so total 12 ways (C)



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Re: A math problem [#permalink]
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24 Jul 2011, 06:55
tracyyahoo wrote: If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2,3,6, what is the greatest number of fourdigit extensions that the company can have?
a)4 b)6 c)12 d)16 e)24
pls help The number has to be even hence the 4 digit number should end with either 2 or 6. Secondly the question states that all the numbers must be used hence repetitions are not allowed. Hence: We can have _ _ _ 2 or _ _ _ 6. So now we need to only fill 3 places with digits 1,6,3 or 1,2,3 respectively Hence 3!*2 = 12. C



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Re: A math problem [#permalink]
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25 Jul 2011, 03:20
Could u explain more details? uote="Sudhanshuacharya"] tracyyahoo wrote: If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2,3,6, what is the greatest number of fourdigit extensions that the company can have?
a)4 b)6 c)12 d)16 e)24
pls help The number has to be even hence the 4 digit number should end with either 2 or 6. Secondly the question states that all the numbers must be used hence repetitions are not allowed. Hence: We can have _ _ _ 2 or _ _ _ 6. So now we need to only fill 3 places with digits 1,6,3 or 1,2,3 respectively Hence 3!*2 = 12. C[/quote] Posted from my mobile device



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Re: A math problem [#permalink]
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25 Jul 2011, 04:08
If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2,3,6, what is the greatest number of fourdigit extensions that the company can have? a)4 b)6 c)12 d)16 e)24 The number for the telephone extension has to be even and 4 digit. Hence the 4 digit number should end with either 2 or 6. Also, the question states that all the numbers must be used because we are talking about greatest number of fourdigit extensions. So, we can have _ _ _ 2 or _ _ _ 6 as four digit even extensions. if the last number is 2 i.e. _ _ _ 2, the blanks can be filled by number 1, 6, 3 and they can be arranged in 3! ways. In the same manner if the last number is 6 i.e. _ _ _ 6, the blanks can be filled by number 1, 2, 3 and they can be arranged in 3! ways. Greatest number of fourdigit extensions = 3!*2 = 12.
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Re: If all of the telephone extensions in a certain company must [#permalink]
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29 Jan 2018, 10:41
Dr_Friedrich wrote: If all of the telephone extensions in a certain company must be even numbers, and if each of the extensions uses all four of the digits 1, 2, 3, and 6, what is the greatest number of fourdigit extensions that the company can have?
(A) 4 (B) 6 (C) 12 (D) 16 (E) 24 Solution: If we let 6 be the last digit, then we have 3! = 6 options for the other 3 digits. If we let 2 be the last digit, we have 6 options also. Therefore, we have a total of 12 options. Answer: C
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